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\begin{document}
\title{Math 525: Lecture 14}
\date{February 22, 2018}
\maketitle
\section{Tightness}
Last lecture, we discussed convergence in distribution, culminating
in Helly's theorem:
\begin{prop}[Helly's theorem]
Let $(F_{n})_{n}$ be a sequence of distribution functions. Then,
there exists a subsequence $(n_{k})_{k}$ and a right continuous nondecreasing
function $F$ such that
\[
F_{n_{k}}(x)\rightarrow F(x)\text{ for all continuity points }x\text{ of }F.
\]
\end{prop}
The issue with the above is that $F$ need not be a distribution function:
\begin{example}
~
\begin{enumerate}
\item Let $X_{n}=n$. Then, $F_{n}(x)=I_{[n,\infty)}(x)$ and $F_{n}(x)\rightarrow0$
for all $x$.
\item Let $X_{n}=-n$. Then, $F_{n}(x)=I_{[-n,\infty)}(x)$ and $F_{n}(x)\rightarrow1$
for all $x$.
\end{enumerate}
Neither $F=0$ nor $F=1$ are distribution functions.
\end{example}
A criteria that ensures the limiting function is indeed a distribution
function is tightness:
\begin{defn}
Let $\{F_{\alpha}\}_{\alpha}$ be a family of distribution functions.
We say $\{F_{\alpha}\}_{\alpha}$ is \emph{tight} if for every $\epsilon>0$,
there exists $r$ sufficiently large such that
\[
F_{\alpha}(r)-F_{\alpha}(-r)\geq1-\epsilon
\]
for all $\alpha$.
\end{defn}
\begin{prop}
Suppose that $(F_{n})_{n}$ is a tight sequence of distribution functions.
Then, there exists a subsequence $(n_{k})_{k}$ and a distribution
function $F$ such that $F_{n_{k}}\Rightarrow F$.
\end{prop}
In other words, the space of distribution functions is \emph{sequentially
compact}.
\begin{proof}
By Helly's theorem, we can find a subsequence $(n_{k})_{k}$ and a
right continuous nondecreasing function $F$ such that
\[
F_{n_{k}}(x)\rightarrow F(x)\text{ for all continuity points }x\text{ of }F.
\]
By tightness, we can find $r$ such that
\[
F_{n}(-r)+\left(1-F_{n}(r)\right)\leq\epsilon.
\]
Since $F_{n}(-r)$ and $1-F_{n}(r)$ are both nonnegative, this implies
\[
F_{n}(-r)\leq\epsilon\qquad\text{and}\qquad1-F_{n}(r)\leq\epsilon.
\]
Now, choose $x_{\epsilon}>r$ so that both $x_{\epsilon}$ and $-x_{\epsilon}$
are continuity points of $F$. Then,
\[
F(-x_{\epsilon})=\lim_{k}F_{n_{k}}(-x_{\epsilon})\leq\epsilon
\]
and
\[
1-F(x_{\epsilon})=\lim_{k}\left\{ 1-F_{n_{k}}(x_{\epsilon})\right\} \leq\epsilon.
\]
Since $\epsilon$ was arbitrary, this implies
\[
\lim_{x\rightarrow-\infty}F(x)=0\qquad\text{and}\qquad\lim_{x\rightarrow\infty}F(x)=1,
\]
as desired.
\end{proof}
\section{Integration}
We will work a lot with integrals today, so let's digress and briefly
talk about integration. Suppose $f\colon\mathbb{R}\rightarrow\mathbb{R}$
is a continuous function. Then, the integral
\[
\int_{a}^{b}f(t)dt
\]
can be interpreted as the limit of Riemann sums.
\begin{figure}[H]
\centering{}\includegraphics[height=3in]{riemann}
\end{figure}
What about when $f$ is not a ``nice'' function? Let $Y\sim U[0,1]$
and define
\[
\int_{a}^{b}f(t)dt\equiv\left(b-a\right)\mathbb{E}\left[f(Y)\right].
\]
The above extends the theory of integration to Borel measurable functions
$f$ (recall that if $f$ is Borel measurable, $f\circ Y$ is a random
variable). You can check that the definition of the integral above
satisfies all the usual conditions (e.g., linearity) and agrees with
the Riemann integral when $f$ is ``nice''.
\begin{rem}
You may have seen the Lebesgue integral. The above is not quite the
Lebesgue integral, since it is only defined for Borel measurable functions
$f$.
\end{rem}
The above tells us that we can apply things like the Monotone Convergence
Theorem, Fatou's Lemma, and the Dominated Convergence Theorem to regular
integrals by treating them like expectations!
\section{LÚvy's continuity theorem}
Our next goal is to establish Paul LÚvy's continuity theorem, which
(roughly speaking) estabishes that a sequence of random variables
converges in distribution if and only if their characteristic functions
converge pointwise.
\subsection{Forward direction}
We have already done all the hard work to prove the forward direction.
\begin{prop}
\label{prop:forward}If $X_{n}\xrightarrow{\mathcal{D}}X$ (i.e.,
$F_{n}\Rightarrow F$) then $\mathbb{E}[e^{itX_{n}}]\equiv\phi_{n}(t)\rightarrow\phi(t)\equiv\mathbb{E}[e^{itX}]$
for all $t$.
\end{prop}
\begin{proof}
Remember that convergence in distribution was equivalent to
\[
\mathbb{E}\left[f(X_{n})\right]\rightarrow\mathbb{E}\left[f(X)\right]
\]
for all bounded and continuous $f$. Let $t$ be arbitrary and take
$f(x)=e^{itx}$ so that
\[
\phi_{n}(t)=\mathbb{E}\left[e^{itX_{n}}\right]\rightarrow\mathbb{E}\left[e^{itX}\right]=\phi(t).\qedhere
\]
\end{proof}
\subsection{Reverse direction}
The remainder of this section is dedicated to proving the reverse
direction. Before we do so, let $X$ be a random variable and $\phi$
its characteristic function. Then, for any $T>0$,
\begin{multline*}
\frac{1}{2T}\int_{-T}^{T}\phi(t)dt=\frac{1}{2T}\int_{-T}^{T}\mathbb{E}\left[e^{itX}\right]dt\\
=\frac{1}{2T}\mathbb{E}\left[\int_{-T}^{T}e^{itX}dt\right]=\frac{1}{2T}\mathbb{E}\left[\frac{2\sin(TX)}{X}\right]=\mathbb{E}\left[\frac{\sin(TX)}{TX}\right]
\end{multline*}
where we have used the Fubini-Tonelli theorem to move the integration
into the expectation (take this as fact if you have yet to encounter
Fubini-Tonelli). Now, let $A>0$ and note that if $|x|>2A$, then
\[
\left|\frac{\sin(Tx)}{Tx}\right|\leq\frac{1}{2TA}.
\]
Let $B=\{-2A0$, there exists $\delta>0$ such that
\[
\frac{1}{\delta}\left|\int_{-\delta}^{\delta}\phi_{n}(t)dt\right|-1\geq1-\epsilon.
\]
\end{prop}
\begin{proof}
Take $A=1/\delta$ in (\ref{eq:tightness_criterion}).
\end{proof}
We are now ready to prove the reverse direction.
\begin{prop}[LÚvy's continuity theorem]
Let $(F_{n})_{n}$ be a sequence of distribution functions and $\phi_{n}$
be the characteristic function of $F_{n}$. Suppose $\phi_{n}\rightarrow\phi$
pointwise for some function $\phi$ which is continuous at the origin
($t=0$). Then, there exists a distribution function $F$ such that
$F_{n}\Rightarrow F$ and $\phi$ is the characteristic function of
$F$.
\end{prop}
Some notes:
\begin{itemize}
\item When we say ``$\phi$ is the characteristic function of $F$'',
we mean that a random variable $X$ associated to $F$ (e.g., take
$X=F^{-1}(Y)$ where $Y\sim U[0,1]$) has characteristic function
$\phi$.
\item That $\phi$ is a characteristic function is one of the results of
the theorem (the limit of characteristic functions need not be a characteristic
function in general).
\end{itemize}
\begin{proof}
Since $\phi$ is continuous at zero, we can choose $\delta$ small
enough so that
\[
\left|\phi(t)-1\right|2\left(1-\frac{\epsilon}{4}\right)=2-\frac{\epsilon}{2}.
\]
Therefore,
\[
2-\frac{\epsilon}{2}